Question

An incident gamma ray photon of wavelength 1.400 x 10–14 m is scattered *** a stationary nucleus. The shift in wavelength of the photon is measured for various scattering angles, and the results are plotted on the graph shown below.

(a) On the graph, sketch a best-fit curve to the data.

In one of the trials, the photon is scattered at an angle of 120° with its original direction.

(b) Calculate the wavelength of this photon after it is scattered *** the nucleus.

(c) Calculate the momentum of this scattered photon.

(d) Calculate the energy that this scattering event imparts to the recoiling nucleus.

Answer 1

(a)

(b) The graph is one of wavelength shift Δλ. Reading 120° from the graph gives a Δλ value of 8x10–17. The new λ of the photon can be found with λ’= λ+Δλ, which gives us λ’ = 1.408x10–14 m

(c) p = h / λ

p = 4.71x10–20 kg-m/*

(d) Using energy conservation. Ephoton(i) = Ephoton(f) + Enucleus … so the nucleus energy is the difference of the photon energy before and after found with hc / λ. ΔE = hc (1/ λf – 1/ λi) = (1.22x10–25) (1 / 1.4x10–14 – 1 / 1.408x10–14) = 8.08 x 10–14 J.